How to prove that $p_{n+1}p_{n+2}p_{n+3}>p_{n+4}p_{n+5}$? $p_n$ is the $n$-th prime number: If $n > 3$ then: 
$$p_{n + 1} p_{n + 2} p_{n + 3} > p_{n + 4} p_{n + 5}$$
I checked the conjecture above true for first fifty million primes.
If the conjecture above is true, it is stronger than Bonse's inequality.

Could you give your remark, reference, or your proof of conjecture above?

 A: From Pierre Dusart's paper arXiv:1002.0442 (page 2) we know that
$$
p_k \le k \left(\log k + \log\log k - 1 + {\log\log k - 2\over \log k}\right) \qquad\mbox{ for } k\ge 688383,
\tag{1}
$$
$$
p_k \ge k \left(\log k + \log\log k - 1 + {\log\log k - 2.1\over \log k}\right) \qquad\mbox{ for } k\ge 3. \ \ \quad
\tag{2}
$$
Substituting the lower bound $(2)$ for primes in the left-hand side, 
and upper bound $(1)$ for primes in the right-hand side, we prove the desired inequality for 
$k\ge 688383$.
For smaller $k$, the inequality can be verified directly by computer.
A: After noting that $2\cdot3\cdot5\lt7\cdot11$ and $3\cdot5\cdot7\lt11\cdot13$, the inequality to prove can be written as
$$p_np_{n+1}p_{n+2}\gt p_{n+3}p_{n+4}\quad\text{for }n\ge3$$
It suffices to check that $5\cdot7\cdot11\gt13\cdot17$ and $7\cdot11\cdot13\gt17\cdot19$ and then show that
$$p_n^3\gt p_{n+4}^2\quad\text{for }n\ge5$$
By Bertrand's Postulate, we have $p_{n+4}\lt2p_{n+3}\lt4p_{n+2}\lt8p_{n+1}\lt16p_n$, hence $p_{n+4}^2\lt256p_n^2$, and so the inequality certainly holds as soon as $p_n\gt256$.  To dispense with the smaller cases, it suffices to note that $256^2\lt41^3$, $53^2\lt19^3$, and $31^2\lt11^3$ (where $53$ is the third prime after $41$ and $31$ is the third prime after $19$).
A: In 2011, Andy Loo proved that there exists a prime $p_k$ between $3n$ and $4n$. Therefore there exists another prime $\ge p_{k+1}$ between $4n$ and $\frac{4}{3}4n<6n$, so that 
$$
p_{k+1} \le \frac{6n}{3n}p_k \le 2p_k \implies p_{k+2}\le 4p_k.
$$
It follows that if $p_{n+1}\ge 17$ then
$$
p_{n+1}\,\frac{p_{n+2}}{p_{n+4}}\, \frac{p_{n+3}}{p_{n+5}}\ge \frac{p_{n+1}}{16}>1.
$$
A: In general the primes are close enough together that the left-hand side will be a bit larger than $p_n^3$ and the right-hand  side will be a bit larger than $p_n^2$.  So if your inequality fails anywhere it will fail for small values of $p_n$ (or, equivalent, for small values of $n$).  Thus we take a two-pronged strategy: first we'll use inequalities that are known to hold for the primes to derive your inequality when $n$ is large, and then we check explicitly when $n$ is small.
In particular we'll use Bertrand's postulate, which states that $p_{n+1} < 2p_n$, for any $n$.  In particular we can turn this around to get $p_{n+1} > p_{n+4}/8, p_{n+2} > p_{n+4}/4, p_{n+3} > p_{n+4}/2$, and so we have
$$ p_{n+1} p_{n+2} p_{n+3} > p_{n+4}^3 / 64. $$
So to show the desired inequality it suffices to show that $p_{n+4}^3 / 64 > p_{n+4} p_{n+5}$, i. e. that $p_{n+4}^2 > 64 p_{n+5}$.  We have $p_{n+4} > p_{n+5}/2$.  Therefore $p_{n+4}^2 > p_{n+5}^2 / 4$.  So if we have $p_{n+5}^2 / 4 > 64 p_{n+5}$, then your inequality holds.  Rearranging gives $p_{n+5} > 256$.  So your inequality holds if $p_{n+5} > 256$.
If $p_{n+5} \le 256$, you can just check directly, as you already have - observe that $5 \times 7 \times 11 > 13 \times 17$, $7 \times 11 \times 13 > 17 \times 19$, and so on up to $229 \times 233 \times 239 > 241 \times 251$.
A: My prove by ideas of Dr. Arthu, Apply Bertrand's postulate
We have:
$P_{n+4}P_{n+5}<2P_{n+4}^2<8P_{n+3}^2$
But $8P_{n+3}^2<P_{n+1}P_{n+2}P_{n+3}$ when $8P_{n+3}<P_{n+1}P_{n+2}$.
We have:
$8P_{n+3}<16P_{n+2}$
When $P_{n+1}>16$, inequality (1) is true, whith $n>6$, we can check direct with $n=4,5$
